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Question

Question: Find the value of \({{\cos }^{-1}}\left( \cos 1540{}^\circ \right)\)....

Find the value of cos1(cos1540){{\cos }^{-1}}\left( \cos 1540{}^\circ \right).

Explanation

Solution

Hint: Simplify cos1540\cos 1540{}^\circ so that you can use the property cos1(cosx)=x{{\cos}^{-1}}\left( \cos x \right)=x. For this, use the property that a trigonometric operation of the form cos(360×n+x)\cos \left( 360{}^\circ \times n+x \right) can be written as cos(x)\cos \left( x \right). Next use the property cos1(cosx)=x{{\cos }^{-1}}\left( \cos x \right)=x on the simplified expression to arrive at the final answer.

Complete step by step solution:
In this question, we need to find the value of cos1(cos1540){{\cos }^{-1}}\left( \cos 1540{}^\circ \right).
We first need to identify that the range of the function cos1(x){{\cos }^{-1}}\left( x \right) is between 00{}^\circ and 180180{}^\circ .
In our question, we are given 15401540{}^\circ which is not in this range. So, we cannot
directly write cos1(cos1540)=1540{{\cos }^{-1}}\left( \cos 1540{}^\circ \right)=1540{}^\circ . So, we need to
simplify 15401540{}^\circ .
To find this value, we will first evaluate cos1540\cos 1540{}^\circ and then we will come to the inverse part.
First, let us simplify cos1540\cos 1540{}^\circ
We can write 1540 as:
1540=360×4+1001540=360\times 4+100
So, we can write cos1540\cos 1540{}^\circ as the following:
cos1540=cos(360×4+100)\cos 1540{}^\circ =\cos \left( 360{}^\circ \times 4+100{}^\circ \right)
Now, we know the property that a trigonometric operation of the form cos(360×n+x)\cos \left( 360{}^\circ \times n+x \right) can be written as cos(x)\cos \left( x \right).
Here, in this question we have n = 4 and x = 100.
Using this property, we can write the above expression as:
cos1540=cos(360×4+100)\cos 1540{}^\circ =\cos \left( 360{}^\circ \times 4+100{}^\circ \right)
cos1540=cos100\cos 1540{}^\circ =\cos 100{}^\circ
Now, we will come to the inverse part.
We know the property that for an angle x, if the measure of angle x is greater than or equal
to 00{}^\circ and less than or equal to 180180{}^\circ , then the expression cos1(cosx){{\cos }^{- 1}}\left( \cos x \right) can be written as x
i.e. cos1(cosx)=x{{\cos }^{-1}}\left( \cos x \right)=x for 0x1800{}^\circ \le x\le 180{}^\circ
Now, since in our question 100100{}^\circ satisfies the condition of being greater than or
equal to 00{}^\circ and less than or equal to 180180{}^\circ , we can use the above
property on it.
We will use this property to calculate cos1(cos1540){{\cos }^{-1}}\left( \cos 1540{}^\circ \right)
cos1(cos1540)=cos1(cos100){{\cos }^{-1}}\left( \cos 1540{}^\circ \right)={{\cos }^{-1}}\left( \cos 100{}^\circ \right)

cos1(cos100)=100{{\cos }^{-1}}\left( \cos 100{}^\circ \right)=100{}^\circ
Hence, cos1(cos1540)=100{{\cos }^{-1}}\left( \cos 1540{}^\circ \right)=100{}^\circ
This is our final answer.

Note: In this question, it is very important to identify that the range of the function cos1(x){{\cos}^{-1}}\left( x \right) is between 00{}^\circ and 180180{}^\circ . In our question, we are given 15401540{}^\circ which is not in this range. So, we cannot directly write cos1(cos1540)=1540{{\cos }^{-1}}\left(\cos 1540{}^\circ \right)=1540{}^\circ . This would be wrong. So, we need to simplify 15401540{}^\circ to a smaller number such that it can be within the range.